Chapter 15 : Magnetic circuits and transformers

Pg: 668 Ex: 15.3

In [1]:
from __future__ import division
from math import pi
M_r=5000#      #relative permeability
R=10*10**-2#
r=2*10**-2#
N=100#      #number of turns
#complex number 'i' is used as a symbol here
I=2*1J#      #here 'i' represents sin(200*pi*t), not as a complex number
M_o=4*pi*10**-7#      #permeability of free space
M=M_r*M_o#      #permeability of the core material
phi=M*N*I*r**2/(2*R)#      #flux
FL=N*phi#      #flux linkages
print " All the values in the textbook are approximated hence the values in this code differ from those of Textbook"
print 'In the below two values,i represents sin(200*pi*t)'      #t-time
print 'flux = j*{:0.3e} webers'.format(phi.imag)
print 'flux linkages = j*{:.2f} weber turns'.format(FL.imag)
#differentiating 'λ' with respect to t
print 'In the below answer, i represents cos(200*pi*t)'
print 'Voltage induced in the coil = {0:.2f}+j*{1:0.2f} volts'.format((FL*200*pi).real,(FL*200*pi).imag)
 All the values in the textbook are approximated hence the values in this code differ from those of Textbook
In the below two values,i represents sin(200*pi*t)
flux = j*2.513e-03 webers
flux linkages = j*0.25 weber turns
In the below answer, i represents cos(200*pi*t)
Voltage induced in the coil = 0.00+j*157.91 volts

Pg: 670 Ex: 15.5

In [2]:
from __future__ import division
from math import pi
M_r=6000#      #relative permeability
M_o=4*pi*10**-7#      #permeability of free space
w_r=3*10**-2#      #width of rectangular cross-section
d_r=2*10**-2#      #depth of rectangular cross-section
N=500#      #number of turns of coil
B_gap=0.25#      #flux density
gap=0.5*10**-2#      #air gap
#centerline of the flux path is a square of side 6cm
l_s=6*10**-2#      #side of square
l_core=4*l_s-gap#      #mean length of the iron core
A_core=w_r*d_r#      #cross-sectional area of the core
M_core=M_r*M_o#      #permeability of core
R_core=l_core/(M_core*A_core)#      #reluctance of the core
A_gap=(d_r+gap)*(w_r+gap)#      #effective area of gap
M_gap=M_o#      #permeability of air(gap)
R_gap=gap/(M_gap*A_gap)#      #reluctance of gap
R=R_gap+R_core#      #total reluctance
phi=B_gap*A_gap#      #flux
F=phi*R#      #magnetomotive force
i=F/N#      #current
print " All the values in the textbook are approximated hence the values in this code differ from those of Textbook"
print 'Current value = %0.2f amperes'%i
 All the values in the textbook are approximated hence the values in this code differ from those of Textbook
Current value = 2.01 amperes

Pg: 672 Ex: 15.6

In [3]:
from __future__ import division
from math import pi
w_core=2*10**-2#      #width
d_core=2*10**-2#      #depth
A_core=w_core*d_core#      #area of core
M_r=1000#      #relative permeability
M_o=4*pi*10**-7#      #permeability of free space
gap_a=1*10**-2#
gap_b=0.5*10**-2#
N=500#      #number of turns of coil
i=2#      #current in the coil
l_c=10*10**-2#      #length for center path
R_c=l_c/(M_r*M_o*A_core)#      #reluctance of center path
#For left side
#taking fringing ino account
A_gap_a=(w_core+gap_a)*(d_core+gap_a)#      #area of gap a
R_gap_a=gap_a/(M_o*A_gap_a)#      #reluctance of gap a
l_s=10*10**-2#      #side of square
l_core_l=3*l_s-gap_a#      #mean length on left side
R_core_l=l_core_l/(M_r*M_o*A_core)#      #reluctance of core
R_L=R_core_l+R_gap_a#      #total reluctance on left side
#For right side
#taking fringing ino account
A_gap_b=(w_core+gap_b)*(d_core+gap_b)#      #area of gap b
R_gap_b=gap_b/(M_o*A_gap_b)#      #reluctance of gap b
l_s=10*10**-2#      #side of square
l_core_r=3*l_s-gap_b#      #mean length on right side
R_core_r=l_core_r/(M_r*M_o*A_core)#      #reluctance of core
R_R=R_core_r+R_gap_b#      #total reluctance on right side
R_T=R_c+1/((1/R_L)+(1/(R_R)))#      #total reluctance
phi_c=N*i/(R_T)#      #flux in the center leg of coil
#by current-division principle
phi_L=phi_c*R_R/(R_L+R_R)#      #left side
phi_R=phi_c*R_L/(R_L+R_R)#      #right side
B_L=phi_L/A_gap_a#      #flux density in gap a
B_R=phi_R/A_gap_b#      #flux density in gap b
print " All the values in the textbook are approximated hence the values in this code differ from those of Textbook"
print 'flux density in gap a = %0.2f tesla'%B_L
print 'flux density in gap b = %0.2f tesla'%B_R
 All the values in the textbook are approximated hence the values in this code differ from those of Textbook
flux density in gap a = 0.11 tesla
flux density in gap b = 0.22 tesla

Pg: 673 Ex: 15.7

In [4]:
from __future__ import division
N=500#      #number of turns of coil
R=4.6*10**6#      #reluctance of the magnetic path from ex15.5
L=N**2/R#      #inductance
print " All the values in the textbook are approximated hence the values in this code differ from those of Textbook"
print 'Inductance of the given coil = %0.2f milli-henry'%(L*10**3)      #milli-10**-3
 All the values in the textbook are approximated hence the values in this code differ from those of Textbook
Inductance of the given coil = 54.35 milli-henry

Pg: 674 Ex: 15.8

In [5]:
from __future__ import division
R=10**7#      #reluctance of core
N_1=100#      #turns for coil 1
N_2=200#      #turns for coil 2
L_1=N_1**2/R#      #self-inductance of coil 1
L_2=N_2**2/R#      #self-inductance of coil 2
#here, complex number i represents i_1 in textbook
phi_1=N_1*1J/R#      #flux produced by i(i_1)
L_21=N_2*phi_1#      #flux linkages of coil 2 from current in coil 1
M=L_21/1J#      #mutual inductance
#milli-(10**-3)
print 'self-inductance of coil 1 =',L_1*10**3,'milli henry'
print 'self-inductance of coil 2 =',L_2*10**3,'in milli henry'
print 'mutual inductance of the coils =',M*10**3, 'milli henry'
self-inductance of coil 1 = 1.0 milli henry
self-inductance of coil 2 = 4.0 in milli henry
mutual inductance of the coils = (2+0j) milli henry

Pg: 675 Ex: 15.9

In [6]:
from __future__ import division
V_s_rms=4700#      #for source
V_L_rms=220#      #load voltage
tr=V_s_rms/V_L_rms#      #turns ratio
print " All the values in the textbook are approximated hence the values in this code differ from those of Textbook"
print 'The required turns ratio N1/N2 = %0.3f'%tr
 All the values in the textbook are approximated hence the values in this code differ from those of Textbook
The required turns ratio N1/N2 = 21.364

Pg: 676 Ex: 15.10

In [7]:
from __future__ import division
V_1_rms=110#
R_L=10#
tr=5#      #turns ratio(N1/N2)
V_2_rms=V_1_rms/tr#      #primary and secondary voltage relation
#a)open switch
print 'OPEN switch'
print 'Primary voltage = %0.2f volts'%V_1_rms
print 'Secondary voltage = %0.2f volts'%V_2_rms
#As switch is open, current in second winding is 0 which implies the current in primary coil to be 0 (ideal transformer condition)
print 'Current in primary and secondary windings =',0,'in amperes'
#b)closed switch
print 'CLOSED switch'
I_2_rms=V_2_rms/R_L#      #ohm's law
I_1_rms=I_2_rms/tr#      #ideal transformer condition
print 'Primary voltage = %0.2f volts'%V_1_rms
print 'Secondary voltage = %0.2f volts'%V_2_rms
print 'Current in primary winding = %0.2f amperes'%I_1_rms
print 'Current in secondary winding = %0.2f amperes'%I_2_rms
OPEN switch
Primary voltage = 110.00 volts
Secondary voltage = 22.00 volts
Current in primary and secondary windings = 0 in amperes
CLOSED switch
Primary voltage = 110.00 volts
Secondary voltage = 22.00 volts
Current in primary winding = 0.44 amperes
Current in secondary winding = 2.20 amperes

Pg: 677 Ex: 15.11

In [8]:
from __future__ import division
from cmath import sin,cos,polar,phase,sqrt
V_s=1000*complex(cos(0),sin(0))#      #source voltage phasor
R_1=10**3#
R_L=10#
Z_L_1=R_L+1J*20#      #impedance
tr=10#      #turns ratio(N1/N2)
Z_L_2=(tr**2)*Z_L_1#      #reflecting Z_L_1 onto primary side
Z_s=R_1+Z_L_2#      #total impedance seen by the source 

Z_s_max = abs(Z_s)
Z_s_phi=phase(Z_s)
#primary quantities
I_1=V_s/Z_s#
I_1_max = abs(I_1)
I_1_phi = phase(I_1)
V_1=I_1*Z_L_2#
V_1_max=abs(V_1)
V_1_phi=phase(V_1)
#using turns ratio to find secondary quantities
I_2=tr*I_1#
I_2_max=abs(I_2)
I_2_phi=phase(I_2)
V_2=V_1/tr#
V_2_max=abs(V_2)
V_2_phi=phase(V_2)
I_2_rms=I_2_max/sqrt(2)#
P_L=(I_2_rms**2)*R_L#      #power to load
print " All the values in the textbook are approximated hence the values in this code differ from those of Textbook"
#we take real parts of angles to take out neglegible and unnecessary imaginary parts(if any are there)
print 'PRIMARY CURRENT:'
print 'peak value = %0.2f amperes'%I_1_max
print 'phase angle = %0.2f degrees'%((I_1_phi*180/pi).real)
print 'PRIMARY VOLTAGE:'
print 'peak value = %0.2f amperes'%(V_1_max)
print 'phase angle = %0.2f degrees'%((V_1_phi*180/pi).real)
print 'SECONDARY CURRENT'
print 'peak value = %0.2f amperes'%I_2_max
print 'phase angle = %0.2f degrees'%((I_2_phi*180/pi).real)
print 'SECONDARY VOLTAGE'
print 'peak value = %0.2f amperes'%(V_2_max)
print 'phase angle = %0.2f degrees'%((V_2_phi*180/pi).real)
print 'power delivered to load = %0.2f watts'%abs(P_L)
 All the values in the textbook are approximated hence the values in this code differ from those of Textbook
PRIMARY CURRENT:
peak value = 0.35 amperes
phase angle = -45.00 degrees
PRIMARY VOLTAGE:
peak value = 790.57 amperes
phase angle = 18.43 degrees
SECONDARY CURRENT
peak value = 3.54 amperes
phase angle = -45.00 degrees
SECONDARY VOLTAGE
peak value = 79.06 amperes
phase angle = 18.43 degrees
power delivered to load = 62.50 watts

Pg: 678 Ex: 15.12

In [9]:
from __future__ import division
from cmath import polar,pi,sin,cos
V_s=1000*complex(cos(0),sin(0))#      #source voltage phasor
R_1=10**3#
tr=10#      #turns ratio(N1/N2)
V_S=V_s/tr#      #reflected voltage
V_S_max=polar(V_S)[0]
V_S_phi=polar(V_S)[1]
R1=R_1/(tr**2)#      #reflected resistance
#we take real parts of angles to take out neglegible and unnecessary imaginary parts(if any are there)
print 'Reflected voltage:'
print 'Peak value = %0.2f volts'%V_S_max
print 'phase angle = %0.2f degrees'%(V_S_phi*180/pi)
print 'Reflected resistance = %0.2f ohms'%R1
Reflected voltage:
Peak value = 100.00 volts
phase angle = 0.00 degrees
Reflected resistance = 10.00 ohms

Pg: 679 Ex: 15.13

In [10]:
from __future__ import division
from cmath import polar,pi,sin,cos,acos

V_L_max=240#
V_L=V_L_max*complex(cos(0),sin(0))#      #load voltage
R_1=3#
R_2=0.03#
R_c=100*10**3#      #core-loss resistance
tr=10#      #turns ratio(N1/N2)
#leakage reactances
Z_1=1J*6.5#
Z_2=1J*0.07#
Z_m=1J*15*10**3#
P_R=20*10**3#      #rated power
I_2_max=P_R/(V_L.real)#
PF=0.8#      #power factor
phi=-acos(PF)#      #-ve for lagging power
I_2=complex(I_2_max*cos(phi),I_2_max*sin(phi))#      #phasor
I_1=I_2/tr#      #primary current
I_1_max=polar(I_1)[0]
I_1_phi =polar(I_1)[1]
V_2=V_L+(R_2+Z_2)*I_2#      #KVL equation
V_1=tr*V_2#
V_s=V_1+(R_1+Z_1)*I_1#      #KVL equation
V_s_max = polar(V_s)[0]
V_s_phi =polar(V_s)[1]#
P_loss=((V_s_max**2)/R_c)+((I_1_max**2)*R_1)+((I_2_max**2)*R_2)#      #power loss in transformer
P_L=V_L*I_2*PF#      #power to load
P_in=P_L+P_loss#      #input power
P_eff=(1-(P_loss/P_in))*100#
#under no-load condtions
I_1=0#
I_2=0#
V_1=V_s_max#
V_no_load=V_1/tr#
PR=((V_no_load-V_L_max)/V_L_max)*100#
print 'Percent regulation = %0.2f'%PR
Percent regulation = 4.51